If the normal stress on section $\mathrm{a}-\mathrm{a}$ is $40 \mathrm{MPa}$, then the normal force component $\mathrm{N}$ can be computed as:

$$\begin{gathered} \mathrm{\sigma }=\frac{\mathrm{N}}{{\mathrm{A}}_{\mathrm{a}-\mathrm{a}}} \therefore \mathrm{N}=\mathrm{\sigma } {\mathrm{A}}_{\mathrm{a}-\mathrm{a} \\ } \\ \mathrm{N}=(40 \mathrm{N}/{\mathrm{mm}}^2)(850.65 {\mathrm{mm}}^2) \\ =34,026 \mathrm{N} \\ =34.026 \mathrm{kN} \end{gathered}$$